Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

思路

1. 和Level Order Traverse一致，但不同的地方在于，隔行的结果需要翻转。比如1，3，5行是从左到右，2，4，6行就是从右到左。
2. 用一个外部变量Depth记录当前结果是数的第几层，当depth为奇数层时，不做任何处理直接向结果中加入layer； 如果为偶数时，则Collections.reverse(layer)，再将其加入到结果中即可。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        
        if (root == null) return result;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        int depth = 0;
        
        while(!queue.isEmpty()) {
            int size = queue.size();
            depth++;
            List<Integer> layer = new ArrayList<Integer>();
            
            for (int i = 0; i < size; i++) {
                TreeNode curNode = queue.poll();
                if (curNode.left != null) {
                    queue.add(curNode.left);
                }
                
                if (curNode.right != null) {
                    queue.add(curNode.right);
                }
                
                layer.add(curNode.val);
            }
            
            if (depth % 2 == 0) {
                Collections.reverse(layer);
            }
            result.add(layer);
        }
        return result;
    }
}